How to Calculate Work Done from an Acceleration vs. Time Graph?

AI Thread Summary
To calculate work done from an acceleration vs. time graph, first convert acceleration values to force using F=ma, where the mass is 2 kg. The area under the force vs. displacement graph represents the work done, which can be calculated by identifying the shape of the area (triangles or rectangles) and applying the appropriate formulas. A common mistake is misreading the graph's scale, which can lead to incorrect calculations; careful attention to graph notations is crucial. Integration can also be used to find velocity and work done, following the work-energy theorem. Understanding the relationship between acceleration, force, and displacement is essential for accurate calculations.
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Homework Statement



Given Data: Graph, 2 kg particle.I am asked to calculate the work done from 0 to 4 meters, or some other interval. Normally I would simply multiply the FORCE of the given particle by multiplying 6m/s^2 times 2kg to create a new force vs. displacement graph, and then calculate the area under the linear piecewise curve. I am doing something wrong

Homework Equations


[/B]
acceleration_vs_time_work_by_falchiongpx-d82wk1e.jpg


The Attempt at a Solution



Calculate the forces as 2kg*acceleration to get the variable force y-values. Calculate area under curve... does not work according to book. Do not seem have enough info for an integral solution.

Sample attempt for Work (J) done from 1 to 2 meters..

1/2(1m*(2kg*18 m/s^2))
 
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You appear to have an acceleration vs displacement graph (bottom horizontal axis).
Work is the area under the force vs displacement graph.
From 1-2m the area is a triangle ... which you did.
What is the problem?
 
In the first meter, judging from the picture, the triangle under the line has an area of (1 m*18 m/s2)/2 = 9 m2/s2

Now you multiply that by the 2kg of mass, and get 18 Joule. The units are right: acceleration [LT-2] by displacement [L] by mass [M], hence ML2T-2.

That way, you can continue the integration 'by eye'...
 
I found out that I misreading the graph from the book. The book labels the scale as = 6m/s2. Every horizontal line represented was actually 2m/s2, with the graph maxing out at 6m/s2. Unfortunately, as was actually the MAX value for the graph. My problem was with reading the scale incorrectly. The teacher was kind enough to go over the problem with me where I realized that each increment of the y-value was 2 m/s2, instead of 6 m/s2.

Note: apologies if I mislabeled the last graph. Please let me know if this is wrong.

Corrected Graph Accel vs Displacement Graph, this is a better representation of what was in the book:

While as = 6m/s

acceleration_vs_disp_work_corrected_by_falchiongpx-d82x92j.jpg

Solution:
a) Read the graph carefully, and locate all notations, even if you require a microscope.
b) Convert y-values on the graph from acceleration to FORCE via F=ma:
6m/s2(2 kg)=12 Newtons, hold the figs

i. Re-graph as a graph of F (x), and calculate the area under the curve for the W.
f_x__vs_disp_for_work_by_falchiongpx-d82x9f4.jpg


ii. For finding the velocity at x meters, take the integral of F(x)...

Antiderivative of F(x) results in 1/2mv2
InteGRAL over a an interVAL.
(I'm new to integration)

W=W=1/2mv2 - 1/2mv20

Which is coincidental:
ΔKE=KE-KE0

And where initial velocity is zero.

v=√[(2*J)/(m)] → v=√[(2*J)/(2 kg)]
​
Does the particle change direction? Answer: what has the greatest energy? The energy for a slowdown or the energy for a speedup?

This allows for solving Work Done, and Velocity from a(x). Acceleration as it relates to Displacement.​
 
Last edited:
ii. For finding the velocity at x meters, take the integral of F(x)...
... or use the work-energy theorem.
 
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